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Iteru [2.4K]
3 years ago
15

The concentration of the particles in the solid phase always remains the same regardless of how much quantity is present true or

false
Chemistry
2 answers:
diamong [38]3 years ago
7 0

Answer: The given statement is false.

Explanation:

Total number of particles present in a substance determines the concentration of particles present.

It is known that in solids, molecules are held together due to strong intermolecular forces of attraction. As a result, solids have definite shape and volume.

So, more is the number of particles present in the solid phase more will be the concentration in solid phase.

Therefore, we can conclude that the statement, concentration of the particles in the solid phase always remains the same regardless of how much quantity is present, is false.

Elan Coil [88]3 years ago
3 0
The answer is true solid phase remains the same
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<u>Answer:</u> The equilibrium constant for the total reaction is 4.09\times 10^{-6}

<u>Explanation:</u>

We are given:

K_{c_1}=0.282\\\\K_{c_2}=41

We are given two intermediate equations:

<u>Equation 1:</u> N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g);K_{c_1}=0.282

The expression of K_{c_1} for the above equation is:

K_{c_1}=\frac{[NH_3]^2}{[N_2][H_2]^3}

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<u>Equation 2:</u> H_2(g)+I_2(g)\rightleftharpoons 2HI(g);K_{c_2}=41

The expression of K_{c_2} for the above equation is:

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Cubing both the sides of equation 2, because we need 3 moles of HI in the main expression if equilibrium constant.

(41)^3=\frac{[HI]^6}{[H_2]^3[I_2]^3}

Now, dividing expression 1 by expression 2, we get:

\frac{K_{c_1}}{K_{c_2}}=\left(\frac{\frac{[NH_3]^2}{[N_2][H_2]^3}}{\frac{[HI]^6}{[H_2]^3[l_2]^3}}\right)\\\\\\\frac{0.282}{68921}=\frac{[NH_3]^2[I_2]^3}{[N_2][HI]^6}

\frac{[NH_3]^2[I_2]^3}{[N_2][HI]^6}=4.09\times 10^{-6}

The above expression is the expression for equilibrium constant of the total equation, which is:

2NH_3(g)+3I_2(g)\rightleftharpoons 6HI(g)+N_2;K_c

Hence, the equilibrium constant for the total reaction is 4.09\times 10^{-6}

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